Name (print, please) _____________________________________________ ID ___________________________
Statistical Quality Control & Design 73-305 Summer 2002
Odette School of Business
University of Windsor
Final Exam Solution
Wednesday, August 21, 8:30 – 11:30 am, ED 1101
Instructor: Mohammed Fazle Baki
Aids Permitted: Text, calculator, straightedge, and three one-sided formula sheets.
Time available: 3 hours
Instructions:
 This exam has 25 pages including this cover page and 7 pages of table.
 Please be sure to put your name and student ID number on each page.
 Show your work.
Marking Scheme:
Question
1
3
5
7
9
11
Score
/15
/3
/6
/12
/6
/14
Question
2
4
6
8
10
Total
Score
/4
/7
/15
/12
/6
/100
Name:______________________________________________
ID:_________________________
Question 1: (15 points)
1.1 Smaller samples are
a. less sensitive to the changes in the quality levels
b. more sensitive to the changes in the quality levels
c. less satisfactory as an indicator of the assignable causes of variation
d. more satisfactory as an indicator of the assignable causes of variation
e. a and c
f. b and c
1.2 For the p -chart with variable sample sizes, the control limits can be calculated using n ave
a. always
b. never
c. whenever individual sample sizes vary more than 25% from the calculated n ave
d. whenever individual sample sizes vary no more than 25% from the calculated n ave
e. all of the above
f. none of the above
1.3 Consider construction of p -chart with variable sample sizes and using n ave .
a. If a sample larger than n ave produces a point inside the upper control limit, the individual
control limit should be calculated to see if the process is out-of-control
b. If a sample larger than n ave produces a point outside the upper control limit, the individual
control limit should be calculated to see if the process is out-of-control
c. If a sample smaller than n ave produces a point inside the upper control limit, the individual
control limit should be calculated to see if the process is out-of-control
d. If a sample smaller than n ave produces a point outside the upper control limit, the individual
control limit should be calculated to see if the process is out-of-control
e. a and d
f. b and c
1.4 Which chart applies to defects and which chart to defectives?
a. The np chart applies to the total number of defects in the samples of constant size
b. The np chart applies to the total number of defectives in the samples of constant size
c. The c chart applies to the total number of defects in the samples of constant size
d. The c chart applies to the total number of defectives in the samples of constant size
e. a and d
f. b and c
1.5 Use the following chart(s) if the number of products produced is too small to form traditional X
and R chart:
a. Individual and moving-range chart
b. Moving-average and moving-range chart
c. Run charts
d. Pre-control charts
e. All of the above
f. None of the above
2
Name:______________________________________________
ID:_________________________
1.6 Consider the problem of economic design of X control chart. If k is decreased keeping
n unchanged
a. Cost of operating in out-of-control state increases
b. Cost of operating in out-of-control state decreases
c. Searching cost increases
d. Searching cost decreases
e. a and d
f. b and c
1.7 Short run charts
a. has no control limits
b. display all the individual values
c. include multiple part numbers on the same chart
d. are used for continuous processes and seasonal products
e. All of the above
f. None of the above
1.8 For which of the following should we use a p control chart to monitor process quality?
a. The volume of scrap parts
b. Weight errors in cans of soup
c. Defective electrical switches
d. Student evaluation on a continuous scale from 1 to 5
e. All of the above
f. None of the above
1.9 You want to determine the lower control line for a p control chart for quality control purposes. You
take several samples of size of 50 items in your production process. From the samples you
determine the fraction defective is 0.002 and the standard deviation is 0.001. If the three-sigma
control limits are calculated, which of the following is the resulting LCL?
a. 0.000
b. 0.002
c. 0.003
d. 0.004
e. 0.005
f. None of the above
1.10 Consider a single sampling plan with sample size n and acceptance number c . The
consumer’s risk increases if
a. only n increases
b. only c increases
c. both n and c increase proportionately
d. a and c
e. all of the above
f. none of the above
3
Name:______________________________________________
ID:_________________________
1.11 Quality control charts usually have a central line and upper and lower control limit lines. Which
of the following are reasons why the process that is being monitored with the chart should be
investigated?
a. Plots fall outside the upper or lower limit lines
b. Normal behavior
c. A large number of plots are on or near the central line
d. No real trend in any direction
e. All of the above
f. None of the above
1.12
a.
b.
c.
d.
e.
f.
What is AOQL?
Proportion of good items in the outgoing lot
Proportion of defective items in the outgoing lot
The maximum proportion of good items in the outgoing lot
The maximum proportion of defective items in the outgoing lot
Proportion of items inspected
Proportion of items replaced by the good items
1.13
a.
b.
c.
d.
e.
f.
For the moving-average and moving-range chart
a run above the central line indicate an out-of-control condition
a run below the central line indicate an out-of-control condition
a point outside control limits indicate an out-of-control condition
a and b
all of the above
none of the above
1.14 You have just used the Capability Index formulas to compute Z(USL)=4.5 and Z(LSL)=3.
Which of the following is the interpretation of these numbers?
a. The capability index, C pk  1.5
b.
c.
d.
e.
f.
The process is not capable
The mean of the production process is not centered
b and c
All of the above
None of the above
1.15 You want to develop a three-sigma X control chart with subgroup size 4. You know that the
standard deviation of the sample means is 10 and the average of the sample means is 300.
Which of the following is the resulting UCL
a. 340
b. 330
c. 315
d. 300
e. 285
f. 270
4
Name:______________________________________________
ID:_________________________
Question 2: (4 points)
An amplifier has a constant failure rate of 20% per 2000 hour.
a. (1 point) What is the probability that the amplifier will survive 8000 hours?
Plife  8000   1     1  0.20   0.4096
4
t
b. (1 point) What is the mean time to failure?

1


1
 5 in 2000 hours = 5 (2000) hours = 10,000 hours
0.20
c. (2 points) What failure rate is required to have a probability of survival of 60% at 8000 hour?
1   4
 0.60
or, 1    0.60 1 / 4
or,   1 - 0.60 1 / 4  0.1199
Hence, the failure rate is 0.119911.99% per 2,000 hour
Question 3: (3 points)
Consider the following system:
A
B
0.80
0.80
0.50
0.50
a. (1 point) Find the reliability of A
RA  0.80  0.80  0.64
b. (1 point) Find the reliability of B
RB  0.50  0.50  0.25
c. (1 point) Find the reliability of the system
Rsystem  1  1  R A 1  RB   1  1  0.64 1  0.25   1  0.36  0.75  0.73
5
Name:______________________________________________
ID:_________________________
Question 4: (7 points)
A paranoid citizen has installed the home-alert system shown below. Find the overall system
reliability.
Camera
Window/door alarms
0.80
0.70
Gate
0.90
Light Beams Dogs
0.80
0.80
0.90
0.80
0.90
0.70
0.70
rgate  0.90  0.801  0.90   0.98
rlight beams  1  1  0.80 1  0.80   0.96
rdogs  1  1  0.90 1  0.90   0.99
rlight beams and dog  0.96  0.99  0.9504
rcamera, light beams and dog  1  1  0.80 1  0.950   0.99008
rwindow/door alram  1  1  0.701  0.701  0.70  0.973
rsystem  0.98  0.99008  0.973  0.9441
Question 5: (6 points)
In an acceptance sampling plan, 20 items were to be tested for 500h with replacement and with an
acceptance number of 2. Answer the following questions using Poisson distribution table.
a. (1 point) If the mean life of items is 2,500 hours, what is the probability that the lot will be
accepted?
T  20500  10,000 hours
1
1
per hour
 
 2,500
1
  T 
10,000  4
2,500
Pa  PNumber of defective  c  2 |   4  0.238
6
Name:______________________________________________
ID:_________________________
b. (3 points) Repeat part a with mean lives 5000, 10000 and 20000 hours.
1

  T
Pa  PNumber of defective  c  2 |   T 


5000
0.0002
10000(0.0002)
Pa  PNumber of defective  c  2 |   2  0.677
=2
10,000
0.0001
10000(0.0001)
Pa  PNumber of defective  c  2 |   1  0.920
=1
20,000
0.00005
20000(0.00001)
Pa  PNumber of defective  c  2 |   0.5  0.986
=0.5
c. (2 points) Plot an Operating Characteristic (OC) curve from the results obtained in parts a and b
Probability of Acceptance
OC Curve
1.2
1
0.8
0.6
0.4
0.2
0
0
5000
10000
15000
20000
25000
Average life
Question 6: (15 points)
A real estate firm evaluates incoming selling agreement forms using the single sampling plan
N  1000 , n  30 and c  1 .
a. (4 points) Compute Pa for p  0.01, 0.02 and 0.03 and using Poisson distribution table.
At p  0.01 , Pa  Pc  1 |  np  np  30  0.01  0.30   0.963
At p  0.02 , Pa  Pc  1 |  np  np  30  0.02  0.60   0.878
At p  0.03 , Pa  Pc  1 |  np  np  30  0.03  0.90   0.772
b. (3 points) Compute Pa for p  0.075 and using Normal approximation.

Pa  P x  1 |   np  30  0.075  2.25,   np1  p   30  0.075  0.925  1.4427
1.5  


 P y  1.5 |   2.25,   1.4427 P z 
|   2.25,   1.4427 



1.5  2.25 

 P z 
  Pz  0.52   0.3015
1.4427 

7

Name:______________________________________________
ID:_________________________
c. (3 points) Compute Pa for p  0.15 and using Binomial formula.
Pa  Px  1 | n  30, p  0.15 
 30 
 30 
30!
0
30
1
29
0.15 0  0.85 30  30! 0.15 1  0.85 29
  0.15   0.85    0.15   0.85  
30! 0!
29! 1!
0
1
 1 1 0.85   30  0.15   0.85   0.0076  0.0404  0.048
d. (2 points) Compute producer’s risk at p  0.01 and consumer’s risk at p  0.15 .
At p  0.01 , producer’s risk = 1- Pa =1-0.963=0.037
30
1
29
At p  0.15 , consumer’s risk = Pa =0.048 (from part c)
or, consumer’s risk = Pa  Pc  1 |  np  np  30  0.15  4.5  0.061 (using Poisson approximation)
e. (2 points) Using results of parts a, b and c, plot an Operating Characteristic (OC) curve.
Probability of
Acceptance, Pa
OC Curve
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.05
0.1
0.15
0.2
Proportion of defective, p
f. (1 point) Using result of part a, compute average outgoing quality for p  0.01
N n
1000  30
Pa p 
0.963  0.01  0.00934  0.934%
At p  0.01, AOQ 
N
1000
Question 7: (12 points)
An item is made in lots of 400 each. The lots are given 100% inspection. The record sheet for the
first 30 lots inspected showed that a total of 240 items did not conform to specifications.
a. (3 points) Determine the centerline and trial control limits for a p chart.
 np  240  0.02
p
 n 400  30




UCL p  p  3
p 1 p
0.021  0.02 
 0.02  3
 0.02  0.021  0.041
n
400
LCL p  p  3
p 1 p
0.021  0.02 
 0.02  3
 0.02  0.021  0.001  0 negative
n
400
8
Name:______________________________________________
ID:_________________________
b. (2 points) If the average fraction of nonconforming items remains unchanged, what is the
probability that the 31st lot will contain 11 or more nonconforming units? Use Poisson distribution
table.




P x  11 | n p  400  0.02  8  1  P x  11 | n p  10  1  0.817  0.183
c. (2 points) If the average fraction of nonconforming items remains unchanged, what is the
probability that the 31st lot will contain 5, 6, 7, 8 or 9 nonconforming units? Use Poisson
distribution table.

 
 

P 5  x  9 | n p  400  0.02  8  P x  9 | n p  10  P x  9 | n p  10  0.718  0.100  0.618
d. (5 points) Among the 30 lots, lot number 5, 16, 21 and 25 contain 13, 23, 14, and 21
nonconforming items respectively. Each of the other lots at the most 10 nonconforming items.
Determine the revised centerline and control limits for the p chart.
Lot 5: np  13, n  400.  p  13 / 400  0.0325  0.041, do not discard
Lot 16: np  23, n  400.  p  23 / 400  0.057  0.041, discard
Lot 21: np  14, n  400.  p  14 / 400  0.035  0.041, do not discard
Lot 25: np  21, n  400.  p  21 / 400  0.0525  0.041, discard
Hence, lots 5 and 21 are in-control and lots 16 and 25 are out of control.
So, discard lots 16 and 25.
 np  npd  240  23  21  196  0.0175
p new 
 n  nd 12000  400  400 11,200




UCL p  p new  3
p new 1  p new
0.01751  0.0175 
 0.0175  3
 0.0175  0.0197  0.0372
n
400
LCL p  p new  3
p new 1  p new
0.01751  0.0175 
 0.0175  3
 0.0175  0.0197  0.0022  0 negative
n
400
Question 8: (12 points)
Consider the following information about mistakes made on tax forms.
Subgroup Number
Number Inspected
Number Nonn
conforming
np
1
40
1
2
40
10
3
40
2
4
60
1
5
60
13
6
60
3
9
Proportion Nonconforming
p
0.025
0.25
0.05
0.01667
0.21667
0.05
Name:______________________________________________
ID:_________________________
a. (4 points) Determine the centerline and trial control limits for a p chart. Since the number
inspected varies, use nave to calculate the centerline and control limits.
1
40  40  40  60  60  60   1 300   50
6
6
 np  1  10  2  1  13  3  30  0.10
p
 n 40  40  40  60  60  60 300
nave 




UCL p  p  3
p 1 p
0.11  0.1
 0.1  3
 0.1  0.1273  0.2273
n
50
LCL p  p  3
p 1 p
0.11  0.1
 0.1  3
 0.1  0.1273  0.0273  0 negative
n
50
b. (2 points) Construct the p chart. Draw the centerline and control limits. Plot all the points.
Proportion of Defective
p Chart
0.3
UCLp
pbar
p
LCLp
0.2
0.1
0
0
2
4
6
8
Subgroup Number
c. (2 points) Identify check points.
Point 2 (outside UCL) and Point 5 (near UCL) are check points.
d. (4 points) For each check point, check if it is needed to compute the individual control limits. If it is
needed to compute the individual control limits, then compute the individual control limits and
comment if the point is in control or out-of-control.
Check Point 2:
 Out of control for n ave
 nind  40  50  nave . So, individual control limits are wider. So, compute individual control
limits.



p 1 p
0.11  0.1
 0.1  3
 0.1  0.1423  0.2423  0.25
n
40
The point is out of control for both n ave and nind
UCL p  p  3
10
Name:______________________________________________
ID:_________________________
Check Point 5:
 In control for n ave
 nind  60  50  nave . So, individual control limits are narrower. So, compute individual
control limits.



p 1 p
0.11  0.1
 0.1  3
 0.1  0.1162  0.2162  0.21667
n
60
The point is out of control (for nind )
UCL p  p  3
Question 9: (6 points)
NB Manufacturing has ordered a new machine. During today’s runoff the following data were
gathered concerning the runout for the diameter of the shaft machine by this piece of equipment. A
precontrol chart was used to set up the machines. Recreate the precontrol chart from the following
data. The tolerance associated with this part is a maximum runout value of 0.0040 (upper
specification). The optimal value is 0.0000 (no runout), the lower specificatin limit.
a. (3 points) Calculate the range values corresponding to each of the red, yellow and green zones.
Red
USL = 0.0040
Yellow
0.0030
Green
0.0020
Green
0.0010
Yellow
LSL = 0.0000
b. (3 points) For each of the following measurements, remark what action will be taken.
Measurement
Action
0.0045
Red, stop, make correction, reset the process
0.0015
Green, continue
0.0035
Yellow, check another piece
0.0005
Yellow (opposite side), stop, reduce variation, reset the process
0.0008
Yellow check another piece
0.0009
Yellow (same side), reset the process
11
Name:______________________________________________
ID:_________________________
Question 10: (6 points)
A quality analyst is checking the process capability associated with the production of struts,
specifically the amount of torque used to tighten a fastener. A total of 30 subgroups each of size 6
have been taken. The upper and lower control limits for the X chart are 76 Nm and 88 Nm,
respectively. X is 82 Nm. R is 5.068 Nm. The specification limits are 80 Nm  10. Calculate
^
a. (1 point) 6 
Since n  6 , d 2  2.534 (see Appendix 2)
^

R 5.068

2
d 2 2.534
^
So, 6  6  2  12
b. (1 point) C p
USL = 80 + 10 = 90 Nm
LSL = 80 - 10 = 70 Nm
USL - LSL 90  70
Cp 

 1.6667
^
12
6
c. (2 points) C pk
z USL  
USL - X
z LSL  
X  LSL
C pk 
^

^


90  82
4
2

82  70
6
2
z min  min 4,6 4

  1.3333
3
3
3
d. (2 points) Interpret the values of C p and C pk

Since C p  1.6667 >1, the variation is small.

Since C pk  1.333  1, the process is capable.

Since C p  C pk , the process is not centerd
So, the process is capable, but needs centering.
12
Name:______________________________________________
ID:_________________________
Question 11: (14 points)
Control charts for X and R are maintained on the shear strength of spot welds. One hundred
observations divided into subgroups of size four are used as a baseline to construct the charts, and
estimates of  and  are computed from these observations. Assume that the 100 observations are
X 1, X 2 ,, X 100 and the ranges of 25 subgroups are R1 , R2 ,, R25 . From these baseline data the
following quantities are computed:
100
 X i  78,000,
i 1
25
R
j 1
j
 1,030
a. (1 point) Verify the estimates that   780 and   20
25
100
X
X
i 1
i
nk
78,000

 780, R 
4  25
R
j 1
k
j

1,030
R
41.2
 41.2,  

 20
25
d 2 2.059
b. (2 points) Find the three-sigma control limits for both X and R charts.
UCLX  X  A2 R  780  0.72941.2  810.0348, LCLX  X  A2 R  780  0.72941.2  749.9652
UCLR  d 4 R  2.28  41.2  94.0184 , LCLR  d 3 R  0  41.2  0
c. (2 points) Type 1 error probability, : For the control limits obtained in part b, what is the
probability of concluding that the process is out of control when it is actually in control?

  2 P X  810 |   780,   20, n  4

810  780 

 2 P z 
  2 Pz  3  21  0.9987   0.0026
20 / 4 

d. (2 points) Suppose that there is a probability of 0.03 that the process shifts from an in-control
state to an out-of-control state in any period. What is the expected number of periods per cycle
that the process remains in control? Assuming each search costs $920, compute the expected
searching cost per cycle.
1 
1  0.03
 32.33

0.03
Expected search cost per cycle  a2 1  E T   9201  0.0026  32.33  $997.33
E T  

e. (2 points) Type 2 error probability, : Assume that when the process shifts out of control, it can be
attributed to a single assignable cause; the magnitude of the shift is about nearly 2  40.
Suppose that the process mean shifts to   2  820 . What is the probability that the X chart
will not indicate an out-of-control condition from a single subgroup sampling?
810  820 
 750  820
  P750  x  810 |   820,   20, n  4   P
z

20 / 4 
 20 / 4
 P 7  z  1  Pz  1  Pz  7  0.1587  0  0.1587
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f. (2 point) What is the expected number of periods per cycle that the process remains in an out-ofcontrol state until a detection is made? If the cost of operating in an out-of-control state is $1515
per period, what is the expected cost of operating in an out-of-control state per cycle?
E S  
1
1

 1.1886 periods
1   1  0.1587
Expected cost of operating in out-of-control state = a3 E S   1,5151.1886   $1,800.73
g. (2 points) What is the expected number of periods per cycle? If the cost of sampling is $3.573 per
unit, what is the expected cost of sampling per cycle?
E C   E T   E S   1.1886  32.3333  32.5219 periods
Expectedsampling cost per cycle  a a nE c   3.573  4  32.5219  $479.09
h. (1 point) Compute the expected total cost of sampling, searching and operating in an out-ofcontrol state per period.
Expectedtotal cost per period 
997.33  1,800.73  479.09 3,277.15

 $97.76 per period
32.5219
32.5219
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